Introduction to Electrodynamics (solutions) - ch06

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Chapter 6
Magnetostatic Fields in Matter
Problem6.1
N B B J.Lo1 [3(
A
)
A
]
A A A A
B
J.Lo ml A=m2 X 1; 1 =_4 "3 mi. r r - ml ; r =y; ml =mlZ; m2 =m2Y' 1 =--4 3z,1rr 1rr
N J.Lomlm2(
A A
)
J.Lomlm2A H 21 b2I S
IN
J.Lo(abI)2A I F. I . .=--4 ~ yxz =--4 ~x. ereml =1ra ,m2= . a =--4 ~x. ma onentatIOn:7rr 7rr r
'downwardI (-z).
Problem6.2
elF=I dl X B; tIN =r X elF =Ir X (dI X B). Now (Prob. 1.6): r X (dI X B) +dl X (B X r) +B X
(r X ill) =O. But d[r X (r X B)] =dr X (r X B) +r X (dr X B) (sinceB is constant),and dr = dI, so
dlX (B X r) = r X (dI X B) - d[r X (r X B)]. Hence2r X (dI X B) =d[r X (r X B)] - B X (r X dI).
dN=!I {d[rX (r X B)] '- B X (r X dI)}.:. N =!I {§d[rX (r X B)] - B X §(r X dI)}.Butthefirstterm
iszero(§d(... ) =0),andthesecondintegralis2a(Eq.1.107).SoN =-I(B X a) =m X B. qed
Problem6.3
(a)
I~ Accordingto Eq.6.2,F = 27rIRBcos().But B =
l!Q.[3(ml.r~r-mtland B cas() = B. YA so B cas() =4". r ' ,
~~[3(ml.i)(i. y) - (mi'y)]. Butml . Y =0 and
i .Y =sinq" while ml . i =ml cos(). :. Bcos() =
~~3ml sinq,casq,.
F =27rIR~~3ml sinq,cosq,.Nowsinq,=~,cosq,=yr2 - R2/r,soF =3~mlIR2~.
ButIR27r= m2, so F = :!.!!!!.2 mlm2~, whileforadipole,R« r, so F = 32
J.LOm1
4m2.". r 1r r
(b)F =V(m2.B) =(m2.V)B = (m2:z)[~z\(~(ml.z)z- ml)]=~mlm2z d~(z\),
~ ~
2ml -3:\z
or,sincez =r: I F =- 3J.Loml m2 A I21r r4 z.
113
114 CHAPTER 6. MAGNETOSTATIC FIELDS IN MATTER
Problem 6.4
dF =J {(dyy) X B(O,y,O)+ (dzz) X B(O,t,z) - (dyy) X B(O,y,t) - (dzz) X B(O,O,z)}
= J {-(dY y) X JB(O,y,t) - B(O,y,On+(dzz) X JB(O,t, z) - B(O,0,zn}
~ y
~ t8B ~ t8B
8z 8y
:::}Jt2
{
zx8
8B_YX8 8
B
}
.
[NotethatJdy~Blo O~t~Bl oooandJdz~B I ~t~B I
.
]Y z z ,y, z, , y O,O,z y 0,0,0
{
X Y z x y Z
}
F =mOO 1 - 0 1 0 =m y8Bx - x8By- x8Bz- z8Bx
8Bz ~!!..!i... 8Bz ~!!..!i... {8y 8y 8z 8z }
8y 8y 8y 8z 8z 8z
[
~ 8Bx ~ 8Bx ~ 8Bx
] (
. . 8By 8Bz 8Bx)=m x-+y-+z- usmgV.B=Otownte-+- =-- .8x 8y 8z 8y 8z 8x
Butm. B =mBx(sincem =mx,here),soV(m. B) =mV(Bx)=m(8fxzx +~y +8fzz z).
ThereforeF =V(m. B). Qed
Problem 6.5 z
(a) B =110Joxy(Prob. 5.14).
m. B =0,soEq.6.3saysI F =0.1
(b) m. B =mWoJox,soIF =mol1oJox.!
y(c) Useproductrule#4: V(p. E)
= p x (V x E) + E x (V x p) + (p . V)E + (E . V)p.
Butp doesnotdependon(x,y,z), sothesecond
andfourthtermsvanish,andV x E =0,sothe
firsttermiszero.HenceV (p.E) =(p . V)E. Qed
Thisargumentdoesnotapplyto themagneticanalog,
sinceV x B =I0.'In fact,V(m. B) =(m. V)B +110(mx J).
(m. V)Ba =mofx(B) =mol1oJoY,(m. V)Bb =mo-/y(l1oJoxy)=O.
Problem 6.6
Aluminum,copper,copperchloride,andsodiumall havean oddnumberofelectrons,soweexpectthemto
beparamagnetic.The rest(havinganevennumber)shouldbediamagnetic.
Problem 6.7
~fi
Jb =VxM =0; Kb =M X ii =M(p.
The field is that of a surfacecurrent Kb =M (P,
but that'sjust a solenoid,sothefield
!outsideis zero,I andinsideB =110Kb =110M.Moreover,it pointsupward(in thedrawing),soI B =110M.!
115
Problem 6.8
VxM =Jb =~!(sks2)z =~(3ks2)Z=3ksz,s vS s Kb =M X Ii =ks2(J>X s) =-kR2z.
Sotheboundcurrentflowsupthecylinder,andreturnsdownthesurface.[Incidentally,thetotalcurrentshould
bezero... is it? Yes,for JJbda = JoR(3ks)(27rsds)= 27rkR3,while JKbdl = (-kR2)(27rR) = -27rkR3.]
Sincethesecurrentshavecylindricalsymmetry,wecangetthefieldby Ampere'slaw:
B .27rS=polenc=Po18 Jbda=27rkpoS3=}I B =POkS2J> I =paM.
Outsidethe cylinder Ienc=0,soI B =0.1
Problem6.9
B
I Kb =M X Ii =M J>.I
(Essentiallya longsolenoid)- --
B (Essentiallya physicaldipole)
(Intermediatecase)
[The externalfields are the sameas in the electrical
case;the internalfields(insidethe bar) are completely
different-in fact,oppositein direction.]
Problem6.10
Kb =M,so the fieldinsidea completering wouldbePoM. The fieldof a squareloop,at the center,is
givenbyProb.5.8:Bsq=V2poI/7rR. HereI =Mw,andR =a/2,so
B - V2poMw - 2V2poMw.
sq- 7r(a/2) - 7ra ' netfieldingap:IB =paM(1- 2:: w) .
116 CHAPTER 6. MAGNETOSTATIC FIELDS IN MATTER
Problem 6.11
As in Sec.4.2.3,wewantthe averageof B = Bout+ Bin' whereBoutis dueto moleculesoutsidea small
spherearoundpointP, andBin is dueto moleculesinsidethesphere.The averageof Boutis sameasfieldat
center(Prob. 5.57b),andfor this it is OK to useEq. 6.10,sincethecenteris "far" fromall themoleculesin
question:
/lo
J
M .
Aout =- ~ d
47f ~2 T
outside
The averageof Bin is '1; (iW )-Eq. 5.89-where m =!7f R3M. ThustheaverageBin is 2/loMj3. But whatis
leftoutof theintegralAoutis thecontributionof a uniformlymagnetizedsphere,to wit: 2/loMj3 (Eq.6.16),
and this is preciselywhat Bin puts back in. So we'll get the correct macroscopicfield using Eq. 6.10. qed
Problem 6.12
(a) M =ksz; Jb =VxM =-kepiKb=M Xft=kRep.
B is in thez direction(thisis essentiallya superpositionof solenoids).So
I B =0 outside.!Usetheamperianloopshown(shaded)-innersideat radiuss:
fB. dl =Bl =/loIenc=/lo[fJbda +Kbl] =/lo[-kl(R - s)+kRl]=/lokls.
:.1B =/lokszinside.I
(b) By symmetry,H pointsin the z direction.That sameamperianloopgivesfH .dl =HI =/loIfenc =0,
sincethereis no freecurrenthere. So I H =0I, and henceI B =/loM.1 OutsideM =0, soB = 0; inside
M =ksz,soB =/loksz.
Problem 6.13
(a) The fieldof a magnetizedsphereis ~/loM (Eq. 6.16),soI B =Bo- ~/lOM,I with thesphereremoved.
In thecavity,H= :oB,soH= :0 (Bo-~/loM) =Ho+M-~M=>IH=Ho+~M.1
(b) ]
K'~
Thefieldinsidea longsolenoidis /loK. HereK =M, sothefieldof theboundcurrenton
the insidesurfaceof thecavityis /loM, pointingdown;Therefore
I B =Bo - /laM; I
H =~(Bo - /laM) =~Bo - M =>I H =Ho.1/lo /lo
(c) (~)Kb This timetheboundcurrentsaresmall,andfar awayfromthecenter,soI B =Bo,I
whileH =:0Bo =Ho+M =>I H =Ho+M.I
[Comment:In the wafer,B is thefieldin themedium;in theneedle,H is the H in themedium;in the
sphere(intermediatecase)bothBand H aremodified.]
117
Problem6.14
M: ~ ; B is thesameasthefieldof a shortsolenoid;H =;0B - M.
Problem6.15
"Potentials":
{
Win(r,O)= EA,r'p'(eosO),(r<R);
Wout(r,O)= E r~lp,(eos0), (r>R).
BoundaryConditions:
{
(i) Win(R,O)=Wout(R,O),
(ii) _8~;utIR+8~nIR=MJ. =Mi. r =Meos0.
(Thecontinuityof W followsfromthegradienttheorem:W(b) - W(a) =J: VW. dl =- J: H.ill;
ifthetwopointsareinfinitesimallyseparated,thislastintegral-+0.)
{
(i) => A,R'= RIf.!.l=>B, =R2'+1A"
(ii) => E(l + 1):.!.2P,(eosO)+ EIA,R'-2P,(eosO) =M cosO.
Combiningthese:
~ - M
L.J(21+I)R' IA,p,(eos0)=M cosO,soA,=0(I '11),and3Al=M =>Al =3.
M M M. 1
ThusWin(r,O)=3reosO=3z, andhenceHin=-V~n =-3z =-3M, so
B =~(H+M)=~ (-~M+M) =1~JLOM.I./
118 CHAPTER 6. MAGNETOSTATIC FIELDS IN MATTER
Problem 6.16
fH. dl =Ilene =I, soH =2~8cP. B =JLo(l+Xm)H = JLo(l+Xm) I A I
Jb =VxM =~~(SXmI )
. 2.,'1>.M ~ XmH ~ I xmJ.1
Sas 211"sz = [QJ Kb = M A
I {
XmI . .2~s </>.
x n = 21TaZ, at
S -- a'
- XmIA '
21Tbz, at r = b.
Total enclosedcurrent,for anamperianloopbetweenthecylinders:
I + X2mI211"a=(1+Xm)I, sof B' dl =JLolenc=JLo(1+ Xm)I::} B = JLo(12+Xm)IcPo,(1I"a 11"S
Problem 6.17
FromEq. 6.20:fH. dl =H(211"s)=Ilene=
{
I(S2/a2), (s<a);
I (s>a).
H =
{
2~:2' (s <a)
}
I {
/La(1+Xm)Is
-L (s» , soB =JLH= 21Ta2,21T8' a I!:SJl.21TS'
(s <a)j
(S> a).
Jb =XmJI (Eq.6.33),andJI = ~, soI Jb =::;I (samedirectionas I).
Kb =M X fi. =XmH X fi. ::} I Kb =;;:I (oppositedirection to I).
h =Jb(ira2)+ Kb(211"a)=XmI - XmI =[Q](asit shouldbe,of course).
Problem 6.18
By themethodof Prob. 6.15:
For larger, we want B(r,O) -+ Bo = Boz, so H = ...LB -+ ...LBoz,and henceW -+ LBoz =/La /La /La
LBorcosO./La
"Potentials":
{
Win(r,O) = L:Alrlpl(cosO), (r<R)j
Wout(r,O)= -~aBorcosO+L:r~l~(cosO),(r>R).
BoundaryConditions:
{
(i) Win(R,O)=Wout(R,O),
(ii) - I/. 8Wout I +I/. 8W;n I - 0,..0 8r R ,.. 8r R - .
(The latter followsfromEq. 6.26.)
(ii) ::} JLo[:0BocosO+2)1 +1):~2 ~(COSO)]+JLLIAIRI-I Pt(cosO)=O.
For 1¥-1, (i) ::}BI = R21+IAI, so [JLo(l+1)+ JLl]AIRI-1 =0,andhenceAI =O.
For1= 1, (i) ::}Al R = L BoR+Bd R2,and(ii) ::}Bo+2JLoBdR3+JLAI = 0,soAl = -3Bo / (2/-LO + /-L)./La
3Bo 3Boz 3Bo A 3Bo
Win(r,O)=-(2 )rcosO=-(2 ). Hin=-VWin= (2 )z= (2 ).JLo+JL JLo+JL /-Lo+JL JLo+/-L
B - H - 3/-LBo-
I (1+Xm)B- /-L - (2JLo+/-L)- 1+Xm/3 o.
119
Bythemethodof Frob.4.23:
Step1: Bo magnetizesthesphere:Mo =XmHo=I'o(~';m)Bo. This magnetizationsets up a field within
thespheregivenby Eq. 6.16:
2 2 Xm 2
B1= -/-LoMo=- Bo=-~Bo
3 3 1+Xm 3
(where~==~ )l+Xm'
Step2: B1 magnetizesthe spherean additional amount M1 =~B1. This setsup an additionalfieldin1'0
thesphere:
2 2
(
2~
)
2
B2 =3"/-LoM1=3"~Bl = 3"' Bo, etc.
The totalfieldis:
2 [ 2 ]
Bo
B =Bo +B1 +B2 +.. . =Bo +(2~/3)Bo+(2~/3)Bo+... = 1+ (2~/3)+ (2~/3) +... Bo =(1 - 2~/3)'
1 - 3 - 3+3Xm - 3(1+Xm)
I
B - ( 1+Xm )B1- 2~/3- 3- 2Xm/(1+Xm)- 3+3Xm- 2Xm- 3+Xm ' so - 1+Xm/3 o.
Problem6.19
~m = -~:::B; M = ~?"= - 4~:~B,whereV is thevolumeperelectron.M =XmH (Eq. 6.29)
2 2
= J.!O(~Xm)B (Eq. 6.30). So Xm =- 4"m:v /-Lo. [Note: Xm « 1, so I won't worry about the (1 + Xm)
term;for thesamereasonweneednot distinguishB fromBelse,aswedid in derivingthe Clausius-Mossotti
equationin Frob. 4.38.] Let's say V = t1l"r3.Then Xm = - ~(4~:r). I'll use 1 A= 10-10m for r.
ThenXm= -(10-7) (4(9~1(~':OX_13~~t:J~10))=1-2X 10-5, I whichis not bad-Table 6.1saysXm=-1 X 10-5.
However,I usedonly oneelectronper atom(copperhas29)anda verycrudevaluefor r. Sincetheorbital
radiusis smallerfor the innerelectrons,they countfor less(D.mrv r2). I havealsoneglectedcompeting
paramagneticeffects.But nevermind. .. this is in therightballpark.
Problem6.20
Placetheobjectin a regionof zeromagneticfield,andheatit abovetheCuriepoint-or simplydropit on
ahardsurface.If it's delicate(awatch,say),placeit betweenthepolesof anelectromagnet,andmagnetizeit
backandforthmanytimes;eachtimeyoureversethedirection,reducethefieldslightly.
Problem6.21
(a)Identicalto Frob. 4.7,onlystartingwith Eqs.6.1and6.3insteadof Eqs.4.4and4.5.
(b)Identicalto Frob. 4.8,but startingwith Eq. 5.87insteadof 3.104.
(c)U =-~~[3COSO1COS02- COS(02- 01)]m1m2.Or, usingCOS(02- Od = COS01COS02- sin01sinO2,
/-Lomlm2
(
. .
)U =-~ smOl sm02- 2COSOlCOS02.411"r
Stablepositionoccursat minimumenergy:g~ =g~=0
{
g~ = 1'°4:~;;'2(cos01sinO2+ 2sin01casO2)=0 =>2sin01cosO2=- cas01sin O2;
g~ = 1'°4:~;;'2(sin01casO2+2cos01sinO2)= 0 =>2sin01cosO2= -4 COS01sinO2,
120 CHAPTER 6. MAGNETOSTATICFIELDS IN MATTER
{
Either sin()1=sin()2=0: ~CD~ or ~@f-
Thus sin ()1cos ()2=sin ()2cos ()1=O. () () 0 t t t Ior cos 1=COS 2 = : or +
@ @
Whichof theseis thestableminimum?Certainlynot@ or@-for thesem2is notparallelto B1, whereaswe
know m2 will line up alongB1. It remainsto compareG)(with ()1=()2=0) and@ (with ()1=rr/2,()2=-rr/2):
U1= /.L°4:~~2(-2); U2= /.L°4:~~2(-1). U1is thelowerenergy,hencethemorestableconfiguration.
Conclusion:They lineup parallel,alongthe linejoiningthem:~ ~, .
(d) They'd line up the sameway: ~ ~ ~ ~ ~ ~
Problem 6.22
F =If dlx B =I (Idl)x Bo+II dlx ((r.Vo)Bo]- I (I dl) x ((ro.Vo)Bo]=II dlx ((r.Vo)Bo]
(becausef dl=0). Now
(dl X BO)i =L Eijkdlj(Boh, and(r. Va) =L rl(Vo)l, so
j,k 1
Fi = I L Eijk [I rl dlj] ((VoMBoh] {Lemma1: 1rl dlj =L Eljmam (proof belOW).}J,k,l m
= I L EijkEljmam(VO)l(Boh
{
Lemma 2: LEijkEljm =Oillhm- OimOkl(proof belOW).
}j,k,l,m j
- I L (OilOkm - OimOkl)am(Vo)l(Bo)k =IL (ak(VoMBoh- ai(Voh(Bo)k]
k,l,m k
= I((Vo)i(a.Bo)-ai(Vo.Bo)].
But V 0.Bo =0 (Eq. 5.48),andm = I a (Eq. 5.84),soF = Vo(m. Bo) (thesubscriptjust remindsusto take
thederivativesat thepointwherem is located). Qed
Proof of Lemma1:
Eq. 1.108saysf(c . r) dl =a x c =-c x a. The jth componentis L:p f cprpdlj =- L:p,mEjpmCpam' Pick
cp=Opl(Le. 1for the lth component,zerofor theothers).Thenf rl dlj =- L:m Ejlmam=L:m Eljmam. Qed
Proof of Lemma2:
EijkEljm =0 unlessijk and ljm arebothpermutationsof 123.In particular,i musteitherbe I or m, andk
mustbetheother,so
L EijkEljm==AOilOkm+ BOimOkl.
j
To determinetheconstantA, picki =I =1,k =m=3;theonlycontributioncomesfromj =2:
E123E123=1=AOllO33 +BO13031 =A =>A =1.
To determineB, picki =m =1,k=I =3:
E123E321= -1 = AO13031+ BOllO33= B =>B = -1.
So
L EijkEljm =OilOkm - OimOklo Qed
j
121
Problem 6.23
(a)Theelectricfieldinsideauniformlypolarizedsphere,E =- 3~0P (Eq.4.14)translatestoH =- 3~0(JLoM) =
-l M. ButB =JLo (H+M). Sothemagneticfieldinsideauniformlymagnetizedsphereis B =JLo(-l M +M) =
I ~/-LOMI (sameasEq. 6.16).
(b)Theelectricfieldinsidea sphereof lineardielectricin anotherwiseuniformelectricfieldis E =1+~e/3Eo
(Eq.4.49).NowXe translatesto Xm,for thenEq. 4.30(P = EoXeE)goesto JLoM = JLoXmH,or M = XmH
(Eq.6.29).So Eq. 4.49=}H = 1+x1m/3Ho.But B ~ JLo(1+ Xm)H, andBo = JLoHo(Eqs. 6.31and6.32),
sothemagneticfieldinsidea sphereof linearmagneticmaterialin an otherwiseuniformmagneticfield is
B 1 Bo ( 1+xm ) I(
.
)
( ) =( /3)' or B = /3 Bo asIn Prob.6.18./-La1+Xm 1+Xm JLo 1+Xm
(c)Theaverageelectricfieldoverasphere,duetochargeswithin,isEave= - 4';'0-h. Let'spretendthecharges
arealldueto thefrozen-inpolarizationofsomemedium(whateverp mightbe,wecansolveV.p = -p to find
theappropriateP). In this casethere are no free charges,and p = IP dT, so Eave= - 4';<0b IP dT, which
translatesto
1 1 ! 1Have= - 47rJLo R3 JLoMdT =- 47rR3m.
ButB =JLo(H+M), soBave= -~p +JLoMave,andMave= !~3' solEave= **, I in agreement
withEq. 5.89.(Wemustassumefor thisargumentthat all thecurrentsarebound,but againit doesn'treally
matter,sincewecanmodelanycurrentconfigurationby an appropriatefrozen-inmagnetization.SeeG. H.
Goedecke,Am. J. Phys. 66, 1010(1998).)
Problem6.24
Eq.2.15: E = p{4';<0Iv #zdT'} (foruniformchargedensity);
Eq. 4.9 : V = p. {4';<0 Iv#zdT'} (for uniform polarization);
Eq.6.11: A = JLo€oMx {4';<0Iv#zdT'} (foruniformmagnetization).
{
Ein = p
~
-t-r) (.Prob.2.12),
Forauniformlychargedsphere(radiusR): . ~o R3 ~)Eout = P 3<0~r (Ex. 2.2).
{
~n = -t-(p. f),
Sothescalarpotentialof a uniformlypolarizedsphereis: V, - 10R3(p . ~ )out - 3<0~ r ,
{
A, = ~(M x r)
andthevectorpotential of a uniformly magnetizedsphereis: A m - ~R3 (M ' ~ )out - 3 ~ X r ,
(confirmingtheresultsof Ex. 4.2andof Exs.6.1and5.11).
Problem6.25
(a)Bl =~~z (Eq.5.86,with()=0).Som2.B1= -~r;:. F =V(m.B) (Eq.6.3)=}F = !1z[-~r;:] z =
3t~~2z. This isthemagneticforceupward(ontheuppermagnet);it balancesthegravitationalforcedownward
(-mdgz):
3JLom2
27rZ4 - mdg =0 =}I z =[
3JLom2
]
1/4
27rmdg .
122 CHAPTER 6. MAGNETOSTATICFIELDS IN MATTERl
I
(b) The middlemagnetis repelledupwardby lowermagnetanddownwardby uppermagnet:
3pom2- 3pom2- md9=O.
21TX4 21Ty4
The top magnetis repelledupwardby middlemagnet,andattracteddownwardby lowermagnet:
3J.Lom2- 3pom2 - md =O.
21Ty4 21T(X+y)4 9
S bt t. . 3/<om2[I I 1 + I ]
. + - 0 I 2 + I - 0 . 2 - I + 1U rac mg. 211" X4- y:r- y:r (:z:+y)4-md9 md9- , or X4- y:r (:z:+y)4- ,so. - W1i'f4 (Z7YW'
Let Q ==xly; then 2 ~ c!:r+ ~. Mathematica givesthe numericalsolution Q =I xly =0.850115...1
Problem 6.26
At theinterface,theperpendicularcomponentofB is continuous(Eq.6.26),andtheparallelcomponentof
H is continuous(Eq. 6.25with Kf =0).SoBt =Bt, H~=H~.ButB =JlH (Eq.6.31),so:1B~ =:2B~,
Nowtan{h=BilIBt,andtan92=B~/Bt,so
tan92 BII B.l. BII Jl2- - ---1.-l - ~ - -
tan91- B2,LBII - BII - JlI1 1
(thesameform,thoughfordifferentreasons,asEq.4.68).
Problem 6.27
In viewofEq.6.33,thereisa bounddipoleat thecenter:mb=Xmm.Sothenetdipolemomentatthe
centeris IIlcenter=m +mb= (1+Xm)m= -/!om.Thisproducesa fieldgivenbyEq.5.87:
Bc~nter= 4Jl 13[3(m.f)f - m].d~pole 1Tr
This accountsfor thefirst termin thefield.The remaindermustbedueto theboundsurfacecurrent(Kb)at
r =R (sincetherecanbe no volumeboundcurrent,accordingto Eq. 6.33).Let us makean educatedguess
(basedeitherontheanswerprovidedor ontheanalogouselectricalProb.4.34)thatthefielddueto thesurface
boundcurrentis (for interiorpoints)oftheformBsurface= Am (Le. a constant,proportionalto m). In thatcurrent
casethemagnetizationwill be:
M =XmH= XmB =X4m 13[3(m .f)f - m]+ XmAm.Jl 7I'r Jl
This will produceboundcurrentsJb =VxM =0,asit should,for0 < r < R (noneedto calculatethis
curl-the secondtermis constant,andthefirstisessentiallythefieldofa dipole,whichweknowis curl-less,
exceptat r =0),and
Kb =M(R) X i = 4;~3(~mX i) + X;A(m X f) =Xmm(- 41T~3+ :)sin9~.
But this is exactlythesurfacecurrentproducedby a spinningsphere:K =o"V =O"wRsin9~, with (O"wR)++
Xmm (~ - 411"~3). So thefieldit produces(forpointsinside)is (Eq. 5.68):
2 2
(
A 1
)Bsurface =-3Jlo(O"wR)=-3JloXmm - - 4 R3 .current Jl 1T
123
Everythingis consistent,therefore,providedA = ~J.toXm(~- 41T~3)'orA (1- ~Xm) = -~~~~7.But
X =(J!:...) -1 soA (1- ~+ ~~) = -~(J..-I-'o)orA (1+~ ) = 2(1-'0-~).A = J!:...2(1-'0-1-')andhencem /JO' 3 3 I-' 3 41TR3, I-' 41TR' 41TR3(21-'0+1-')'
J.t
{
I -- 2(J.to- J.t)m
}B= - -d3(m.r)r-m]+ R3(2 )' Qed411"r J.to+ J.t
The exteriorfield is that of the centraldipole plus that of the surfacecurrent,which, accordingto Prob. 5.36,
isalsoa perfectdipolefield,of dipolemoment
msurface= ~1I"R3((Jc..JR)= ~1I"R3(~Bsurface )=211"R3..!!:..-2(J.to - J.t)m =J1(J.to-- J.t)m.current 3 3 2J.to current J.to 411"R3(2J.to+J.t) J.to(2J.to+J.t)
Sothetotaldipolemomentis:
J.t J.t (J.to- J.t) 3J.tm
mtot =-m+ -m =-,
J.to J.to (2J.to+J.t) (2J.to+J.t)
andhencethefield(forr > R) is
B =J.to( 3J.t )
1
411"2J.to+ J.t r3 [3(m. f)f - m].
Problem6.28
Theproblemis that thefieldinsidea cavityis not thesameasthefieldin thematerialitself.
(a)Amperetype.The fielddeepinsidethemagnetis thatof a longsolenoid,Bo ~ J.toM.FromProb.6.13:
{
Sphere: B =Bo -- ~J.toM=iJ.toM;
Needle: B =Bo - J.toM=0;
Wafer: B =J.toM.
(b)Gilberttype.This is analogousto theelectriccase.Thefieldat thecenteris approximatelythatmidway
betweentwodistantpointcharges,Bo ~ O.FromProb.4.16(withE --tB, 1/Eo--t J.to,P --tM):
{
Sphere:B =Bo+'M =iJ.toM;
Needle:B =Bo=0;
Wafer: B =Bo+J.toM=J.toM.
In thecavities,then,thefieldsarethesamefor thetwomodels,andthiswill beno testat all. I Yes.I Fundit
with$1M fromtheOfficeof AlternativeMedicine.